Simplify; express your answer in exponential form. Assume $x\neq 0, n\neq 0$. $\dfrac{{(x^{4})^{-2}}}{{(x^{-2}n^{2})^{4}}}$
Solution: To start, try working on the numerator and the denominator independently. In the numerator, we have ${x^{4}}$ to the exponent ${-2}$ . Now ${4 \times -2 = -8}$ , so ${(x^{4})^{-2} = x^{-8}}$ In the denominator, we can use the distributive property of exponents. ${(x^{-2}n^{2})^{4} = (x^{-2})^{4}(n^{2})^{4}}$ Simplify using the same method from the numerator and put the entire equation together. $\dfrac{{(x^{4})^{-2}}}{{(x^{-2}n^{2})^{4}}} = \dfrac{{x^{-8}}}{{x^{-8}n^{8}}}$ Break up the equation by variable and simplify. $\dfrac{{x^{-8}}}{{x^{-8}n^{8}}} = \dfrac{{x^{-8}}}{{x^{-8}}} \cdot \dfrac{{1}}{{n^{8}}} = x^{{-8} - {(-8)}} \cdot n^{- {8}} = n^{-8}$.